How do you solve $\sqrt{x + 1} = 2$ $sqrt(x+1) = 2$ ?

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How do you solve $\sqrt{x + 1} = 2$ $sqrt(x+1) = 2$ ?

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Answer 1 · 2016-06-02T12:49:11

x = 3

Explanation:

To 'undo' the square root we have to perform the inverse operation.
The inverse to 'square root' is 'square'. Since this is an equation we must square both sides.

$\Rightarrow {(\sqrt{x + 1})}^{2} = 2^{2} \Rightarrow x + 1 = 4 \Rightarrow x = 4 - 1 = 3$ $rArr(sqrt(x+1))^2=2^2rArrx+1=4rArrx=4-1=3$

Answer 2 · 2016-06-02T12:49:50

$x = 3$ $x = 3$

Explanation:

When $\sqrt{x + 1} = 2$ $sqrt(x + 1) = 2$
Square both sides
$x + 1 = 4$ $x + 1 = 4$
Subtract $1$ $1$ from both sides
$x = 3$ $x = 3$

Check when $x = 3$ $x = 3$
$3 + 1 = 4$ $3 + 1 = 4$ and $\sqrt{4} = 2$ $sqrt 4 = 2$
So answer is correct

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How do you solve $\sqrt{x + 1} = 2$ $sqrt(x+1) = 2$ ?

2 Answers

Explanation:

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How do you solve sqrt(x+1) = 2√x+1=2sqrt(x+1) = 2?

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How do you solve $\sqrt{x + 1} = 2$ $sqrt(x+1) = 2$ ?